Answer
$$
\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x =1 .
$$
So, the series converges
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}= \frac{\ln 1}{1^{3}}+ \frac{\ln 2}{2^{3}}+\frac{\ln 3}{3^{3}}+...
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}
$$
The function
$$
f(x)=\frac{\ln x}{x^{2}}
$$
is positive and continuous for $x\gt 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{1-2\ln x}{x^{3}}.
$$
So, $f^{'}(x) \lt 0 $ for $x\gt e^{\frac{1}{2}}$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{\ln x}{x^{2}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=\ln x, \quad\quad dv= \frac{dx}{ x^{2} } }\\ {d u= \frac{dx}{x }, \quad\quad v= - \frac{1}{x} }\end{array}\right] , \text { then }\\
&=\lim\limits_{t \to \infty } [-\frac{\ln x+1}{x}]_{1}^{t} \\
&=\lim\limits_{t \to \infty } [-\frac{\ln t+1}{t}]+1\\
& \quad\quad\quad\left[\text { by using L'Hospital's rule as follows } \right] \\
& =\lim\limits_{t \to \infty } [-\frac{\frac{1}{t}}{1}]+1 \\
&= \lim\limits_{t \to \infty } [-\frac{1}{t}]+1 \\
&=1.
\end{aligned}
$$
So, the series converges